Proof verification and help for the set $[0,1]\times\{0\} \subseteq\mathbb{R^2}$ being closed.

Question

May I please receive feedback and help on this proof? Thank you!

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$[0,1]$ is closed in $\R$. I want to prove that, when considered as a subset of $\R^2$, that is, as a line segment on the $x$-axis in the plane, it is also closed. Specifically, I want to show that the set $[0,1]\times\{0\} \subseteq\R^2$ is closed as well.

$\textbf{Solutiion:}$ Let's consider a point $x=(a_1, b_1) \notin [0,1]\times\{0\}$. Then, either $a_1 \notin [0,1]$ or $b_1 \ne 0$.

$\textbf{Case I:}$ $a_1 \notin [0,1]$

Let $a_1 > 1$. Then choose $\epsilon = \frac{a_1 -1}{2} > 0.$ Then the neighborhood $N_{\epsilon}(x)$ does not contain any point of $[0,1]\times\{0\}$.

Therefore, $x$ is not a limit point of $[0,1]\times\{0\}$. Thus, $[0,1]\times\{0\}$ contains all it's limit points.

$\textbf{Case II:}$ If $b_1 \ne 0$. Then $|b_1| > 0$. Let's choose $\frac{\epsilon}{2}>0.$ Then the $\epsilon$-neighborhood $N_{\epsilon} (x)$ of $x$ contains no point of $[0,1]\times\{0\}$.

Hence, from Case I and Case II, we conclude that $[0,1]\times\{0\}$ contains all of its limit points. Hence $[0,1]\times\{0\}$ is a closed subset of $\R^2$.

Answer 1

In a product topology (as $\Bbb R^2$ also is, $\Bbb R^2 \simeq \Bbb R \times \Bbb R$ topologically) $C_1 \times C_2$ is closed iff $C_1$ and $C_2$ are closed in $\Bbb R$. And $[0,1], \{0\}$ are well-known to be closed in $\Bbb R$ (e.g. their complements are (unions of) open intervals).

Answer 2

You forgot the case $a_1 <0$ in CASE I. In this case choose $\epsilon <|a_n|$.

When $b_1 \neq 0$ you have to choose $\epsilon <|b_1|$.